Answer
$$\frac{1}{16} \sec ^{-1} \frac{z}{2}+\frac{\sqrt{z^{2}-4}}{8 z^{2}}+C$$
Work Step by Step
Given $$\int \frac{d z}{z^{3} \sqrt{z^{2}-4}}$$
Let
$$z=2\sec u\ \ \ \ \ \to\ \ \ \ dz= 2\sec u\tan udu $$
Then
\begin{align*}
\int \frac{d z}{z^{3} \sqrt{z^{2}-4}}&=\int \frac{ 2\sec u\tan udu }{8\sec^3 u \sqrt{4\sec^{2}u-4}}\\
&=\int \frac{\tan u d u}{4 \sec ^{2}u \sqrt{4\left(\sec ^{2}u-1\right)}}\\
&=\int \frac{\tan u d u}{8 \sec ^{2} u \sqrt{\tan ^{2} u}}\\
&=\frac{1}{8} \int \frac{d u}{\sec ^{2} u}\\
&=\frac{1}{8} \int \cos ^{2} u d u\\
&=\frac{1}{16} \int (1+\cos 2 u)du\\
&=\frac{1}{16}\left(u+\frac{\sin 2u}{2}\right)+C\\
&= \frac{1}{16}\left(u+ \sin u\cos u\right)+C\\
&=\frac{1}{16}\left[\sec ^{-1} \frac{z}{2}+\frac{\sqrt{z^{2}-4}}{z} \cdot \frac{2}{z}\right]+C\\
&=\frac{1}{16} \sec ^{-1} \frac{z}{2}+\frac{\sqrt{z^{2}-4}}{8 z^{2}}+C
\end{align*}
